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a) $E^2 - B^2$ is a Lorentz invariant scalar, since it is proportional to the product $F_{\mu \nu} F^{\mu \nu}$ and $F_{\mu \nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$ is built out of 4-vectors and so must itself be Lorentz invariant.

Now if in some frame $\vec B = 0$, then $B^2 = 0$ and $E^2 - B^2 \leq 0$. Conversely if in another frame $\vec E = 0$, then $B^2 = 0$ and $E^2 - B^2 \geq 0$. Now if both conditions hold simultaneously then $$E^2 - B^2 \leq 0 \leq E^2 - B^2 \Rightarrow E^2 = B^2.$$ We have already said that $\vec E = 0$ in some frame, so in that frame $\vec B = 0$ also. But we have shown that for the given conditions, equality of $E^2$ and $B^2$ holds in all frames, i.e. $\vec E = \vec B = 0$ in all frames as required.

b) 


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